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I have a simple linear operator:

$$\begin{align}g: \Bbb{R^4} &\to \Bbb{R^3}\\g (x, y, u , v) &= ( x + u, x + v, y + u)\end{align}$$

How would I determine the image of this linear operator?

I thought that putting in the vectors

$(0, 0, 0, 1) \implies \dots$
$(0, 0, 1, 0) \implies \dots$
$\;\;\;\vdots$

would yield the $4$ vectors of the image.

But the solution says there are only $3$ vectors in the image of this operator.

What am I missing?

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Since the range of the operator is $\mathbb{R}^3$, there can only be $3$ linearly independent vectors in the image. –  robjohn May 4 '12 at 18:24

1 Answer 1

up vote 1 down vote accepted

Firstly, sanity check:

  • Image is a subspace of $\Bbb R^3$ and so it cannot have four vectors. I assume that you're saying that you get $4$ linearly independent vectors, even then, it clearly fails to make sense as the dimension of the image space is atmost $3$.

So, you're not quite right. Notice that the image is the column space of the matrix of transformation (compute that!) , but clearly, $$\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix},\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 0 \\ 1\end{bmatrix},\begin{bmatrix} 0\\ 1 \\ 0\end{bmatrix}$$ are not linearly independent.

Now, you need to compute a basis for the image. Since the four column vectors there span the image, and already $(1,1,0)$ is a linear combination of the others, you can drop that and still the span of the rest of the vectors would be the image. But, the other three are linearly independent. So, they form a basis for your image.

In general, the basis for the image space can be obtained from columns of the matrix in the row-reduced echelon form.

$$\begin{bmatrix} 1&0&1&0 \\ 1&0&0&1 \\ 0&1&1&0\end{bmatrix} \sim \begin{bmatrix} 1&0&1&0 \\ 0&0&-1&1\\ 0&1&1&0\end{bmatrix}$$

From here, it is obvious that $C_3=C_1+C_2-C_4$ and that $C_1, C_2$ and $C_4$ are linearly independent. Thus, $\mathcal{B}= \{C_1, C_2, C_4\}$ form a basis for the image; indeed $\mathcal{B}$ is the standard basis for $\Bbb{R^3}$.

Notice further that the basis we have obtained are in fact different from the one we obtained before.

For more on finding a basis for Column space, refer to: Wikipedia

To check your matrix of transformation, hover below:

The matrix of the linear transformation is: $m(g)=\begin{bmatrix}1 &0&1&0\\1&0&0&1\\0&1&1&0 \end{bmatrix}$

Additional related exercises:

  • See my answer here [It might be nice to spend time to see why the linked exercise actually is related to my observations here.]
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I see. But we don't need the vector (1, 1, 0) in the image because it can be represented as a linear combination of the rest 3. Is this correct? –  Tool May 4 '12 at 17:02
1  
Yes, right. But, you need to compute a basis for the image. Since the four column vectors there span the image, and already (1,1,0) is a linear combination of the others, you can drop that and still the span of the rest of the vectors would be the image. But, the other three are linearly independent. So, they form a basis for your image. –  user21436 May 4 '12 at 17:05
    
Great! One more question - is there an efficient way to quickly spot which of the 4 vectors don't belong in the base? In other words, which vector/s I must eliminate to get the base (I mean how to compute this). I am guessing it has something to do with Gauss-Jordan. –  Tool May 4 '12 at 17:06
    
Yes, the column vectors in the echelon form of your matrix span the image space. –  user21436 May 4 '12 at 17:09
    
I'm not sure if I got this right. I reduced the linear transformation matrix - m(g) and didn't get the 3 base vectors. Used a calculator over here to confirm: math.purdue.edu/~dvb/matrix.html –  Tool May 4 '12 at 17:17

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